Using the trigonometric identity:
#1+tan^2alpha = 1+sin^2alpha/cos^2alpha = (cos^2alpha+sin^2alpha)/cos^2alpha = sec^2alpha#
At the numerator let:
#sec^4x = sec^2x sec^2x = (1+tan^2x)sec^2x#
Then substitute #t= tanx#, #dt = sec^2xdx#:
#int ((4+5tanx)sec^4x)/tan^5x dx = int ((4+5t)(1+t^2))/t^5dt#
#int ((4+5tanx)sec^4x)/tan^5x dx = int (4+5t+4t^2+5t^3)/t^5dt#
and using the linearity of the integral:
#int ((4+5tanx)sec^4x)/tan^5x dx = 4 int (dt)/t^5 +5 int (dt)/t^4+4int (dt)/t^3+5int (dt)/t^2#
#int ((4+5tanx)sec^4x)/tan^5x dx = -1/t^4-5/(3t^3)-2/t^2-5/t+C#
#int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5t+6t^2+15t^3)/(3t^4)+C#
and undoing the substitution:
#int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5tanx+6tan^2x+15tan^3x)/(3tan^4x)+C#
